Form the pde by eliminating the arbitary function z=y²+2f[(1/x)+logy]

Given the equation \( z = y^2 + 2f\left(\frac{1}{x} + \log y\right) \), we need to eliminate the arbitrary function \( f \) to form a partial differential equation (PDE).

### Step 1: Differentiate with respect to \( x \)
First, differentiate the given equation with respect to \( x \):

\[
\frac{\partial z}{\partial x} = 2f'\left(\frac{1}{x} + \log y\right) \cdot \left(-\frac{1}{x^2}\right)
\]

This simplifies to:

\[
\frac{\partial z}{\partial x} = -\frac{2f'\left(\frac{1}{x} + \log y\right)}{x^2}
\]

### Step 2: Differentiate with respect to \( y \)
Next, differentiate the given equation with respect to \( y \):

\[
\frac{\partial z}{\partial y} = 2y + 2f'\left(\frac{1}{x} + \log y\right) \cdot \frac{1}{y}
\]

This simplifies to:

\[
\frac{\partial z}{\partial y} = 2y + \frac{2f'\left(\frac{1}{x} + \log y\right)}{y}
\]

### Step 3: Differentiate \(\frac{\partial z}{\partial y}\) with respect to \( x \)
Differentiate \( \frac{\partial z}{\partial y} \) with respect to \( x \):

\[
\frac{\partial^2 z}{\partial x \partial y} = 2f''\left(\frac{1}{x} + \log y\right) \cdot \left(-\frac{1}{x^2}\right) \cdot \frac{1}{y}
\]

This can be written as:

\[
\frac{\partial^2 z}{\partial x \partial y} = -\frac{2f''\left(\frac{1}{x} + \log y\right)}{x^2 y}
\]

### Step 4: Differentiate \(\frac{\partial z}{\partial x}\) with respect to \( y \)
Differentiate \( \frac{\partial z}{\partial x} \) with respect to \( y \):

\[
\frac{\partial^2 z}{\partial y \partial x} = -\frac{2f''\left(\frac{1}{x} + \log y\right)}{x^2 y}
\]

### Step 5: Combine the results
Since the mixed partial derivatives should be equal, we get:

\[
\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial^2 z}{\partial y \partial x}
\]

This equality confirms that the elimination of the arbitrary function \( f \) has been correctly handled, and thus the PDE formed is:

\[
\frac{\partial^2 z}{\partial x \partial y} = -\frac{2f''\left(\frac{1}{x} + \log y\right)}{x^2 y}
\]

Given that this equation must hold for all \( x \) and \( y \), we conclude that:

\[
\boxed{\frac{\partial^2 z}{\partial x \partial y} + \frac{2}{x^2 y}\frac{\partial^2 z}{\partial y \partial x} = 0}
\]

This is the partial differential equation obtained by eliminating the arbitrary function \( f \).

Would you like more details or have any questions? 

Here are five questions related to this problem:
1. How do you derive the second-order partial derivatives from the original equation?
2. What is the significance of mixed partial derivatives in forming PDEs?
3. Can the method used here apply to other arbitrary functions in different forms?
4. How would the PDE change if the function \( f \) depended on \( x \) or \( y \) separately instead of a combined term?
5. What role does the logarithmic function play in the differentiation process?

**Tip:** Always check that mixed partial derivatives are consistent to ensure the validity of your PDE derivation.

Learn how to eliminate the arbitrary function \( f \) from the given equation \( z = y^2 + 2f\left(\frac{1}{x} + \log y\right) \) to derive a partial differential equation (PDE). This solution involves differentiating with respect to \( x \) and \( y \), and ensuring consistency of mixed partial derivatives. Suitable for advanced undergraduate level students and mathematicians.

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